Different Determinants via Elimination : Add a multiple of one row to another row. [GStrang P247] For simplicity, I consider a general $2$ by $2$ matrix that I call $M.$
$ M:=\begin{bmatrix}
    a & b \\
    c & d \\
    \end{bmatrix} \overset{\frac{-b}{d}R2 + R1 \rightarrow R1}{\longrightarrow}
\begin{bmatrix}
    a - \frac{b}{d}c& 0 \\
    \color{#B8860B}c & d \\
    \end{bmatrix} \overset{-\dfrac{\color{#B8860B}c}{a - \frac{b}{d}c}R1 + R2 \rightarrow R2}{\longrightarrow}
     \begin{bmatrix}
    a - \frac{b}{d}c& 0 \\
    0 & d \\
    \end{bmatrix} \quad (\checkmark)$
$ \begin{bmatrix}
    a & b \\
    c & d \\
    \end{bmatrix} \overset{\frac{-d}{b}R1 + R2 \rightarrow R1}{\longrightarrow}
\begin{bmatrix}
    c - \frac{d}{b}a& 0 \\
    \color{#B8860B}c & d \\
    \end{bmatrix} \overset{-\dfrac{\color{#B8860B}c}{c - \frac{d}{b}a}R1 + R2 \rightarrow R2}{\longrightarrow} \begin{bmatrix}
    c - \frac{d}{b}a & 0 \\
    0 & d \\
    \end{bmatrix} \quad (\yen) $
Both $\frac{-b}{d}R2 + R1 \rightarrow R1$ and $\frac{-d}{b}R1 + R2 \rightarrow R1$ zero the (1, 2) entry.
The middle matrix in $(\checkmark) \implies \det M = ad - bc$
but the middle matrix in $(\yen) \implies  \det M = cd - \frac{d^2}{b}a$.
What's the problem? What can be concluded for the general $n \times n$ case?
(The det can already be espied from the middle matrix; I include the diagonal matrices for completeness)
 A: Recall that performing the elementary row operation of row replacement (adding a multiple of one row, say $k(R_j)$, to another row, say $R_i$) does not change the determinant of a matrix. This certainly holds true for the first checkmarked matrix. The reason why it doesn't work for the second matrix is because you did not perform a row replacement. That is, row replacements have the form:
$$
R_i + k(R_j) \to R_i
$$
whereas you performed an operation of the form:
$$
R_i + k(R_j) \to R_j
$$
A: The following is predicated on Pete L. Clark's comment dated Dec 5 2013. I recast it here because the $i$ and $j$ in Adriano's answer are reversed in Pete L. Clark's comment on Dec 5 2013. I follow Adriano's answer so in what follows, $i$ and $j$ denote the $i$ and $j$ in Adriano's answer. 
It is not a "Type III row operation", where you add a multiple of row $j$ to row $i$ and use that to replace row $i$. 
Per contra, your operation is a composition of the operation of scaling row $j$ by $\color{#FF4F00}k$ and adding the subsequent $\color{#FF4F00}k \times$ row $j$ to row $j$. Thus it has the effect of multiplying the determinant by $\color{#FF4F00}k$. 
Your operation does preserve the row space, so could be taken as a row operation if you wanted; it just affects the determinant differently. You might want to write down the matrix corresponding to your row operation to see what the difference is.
